You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Circular Kinematics

Magnetic fiel...

Question

Magnetic field of magnitude $B_{0}$ exist in the region above the line as shown in figure. A charged particle of mass m and charge q moving at speed 'u' enters magnetic field region. Find time interval for which particle remains in magnetic field.

\frac{2 π m θ}{q B π}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2 π m}{q B} (1 - \frac{θ}{π})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{m θ}{q B}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2 m θ}{q B} (\frac{1}{θ} - \frac{1}{π})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{2 π m}{q B} (1 - \frac{θ}{π})$
By geometry, $α = 2 θ$
Here, $T = \frac{2 π R}{V} = \frac{2 π m}{q B}$
So time $t = \frac{T}{2 π} \times (2 π - α)$
$t = T (1 - \frac{α}{2 π}) = T (1 - \frac{θ}{π})$
$= \frac{2 π m}{q B} (1 - \frac{θ}{π})$

Suggest Corrections

Similar questions

Q. Magnetic field of magnitude

B_{0}

exist in the region above the line as shown in figure. A charged particle of mass m and charge q moving at speed 'u' enters magnetic field region. Find time interval for which particle remains in magnetic field.

Q. A particle moving with velocity

v

having specific charge

(\frac{q}{m})

enters a region of magnetic field

B

having width

d = \frac{3 m v}{5 q B}

at angle

53^{o}

to the boundary of magnetic field. Find the angle

θ

in the shown figure.

Q. A positively charged particle having charge

q

and mass

m

moving with velocity

v_{0}^i

enters a region in which magnetic field

\to B = - B_{0}^k

exists. The magnetic field region extends from

x = 0

x = d

. Then the time spent by the particle in magnetic field if

d

is equal to

\frac{1.5 m v}{q B}

Q. In a certain region uniform electric field

\to E

and uniform magnetic field

\to B

are present in opposite directions. A particle of mass

m

and of charge

q

enters in this region with a velocity

v

at an angle

θ

from the magnetic field. The time after which the speed of the particle would be minimum is equal to :

Q. In region

x > 0

, a uniform and constant magnetic field

_{1} = 2 B_{0} \to k

exists. Another uniform and constant magnetic field

_{2} = B_{0}^k

, exists in region

x < 0

. A positively charged particle of mass

m

and charge

q

is crossing origin at time

t = 0

with a velocity

\to u = u_{0}^i

. The particle comes back to its initial position after a time

(B_{0}, u_{0}

are positive constants)

Join BYJU'S Learning Program

Magnetic field of magnitude B0 exist in the region above the line as shown in figure. A charged particle of mass m and charge q moving at speed 'u' enters magnetic field region. Find time interval for which particle remains in magnetic field.

The correct option is B 2πmqB(1−θπ)By geometry, α=2θ Here, T=2πRV=2πmqB So time t=T2π×(2π−α) t=T(1−α2π)=T(1−θπ) =2πmqB(1−θπ)

Magnetic field of magnitude $B_{0}$ exist in the region above the line as shown in figure. A charged particle of mass m and charge q moving at speed 'u' enters magnetic field region. Find time interval for which particle remains in magnetic field.

The correct option is B $\frac{2 π m}{q B} (1 - \frac{θ}{π})$
By geometry, $α = 2 θ$
Here, $T = \frac{2 π R}{V} = \frac{2 π m}{q B}$
So time $t = \frac{T}{2 π} \times (2 π - α)$
$t = T (1 - \frac{α}{2 π}) = T (1 - \frac{θ}{π})$
$= \frac{2 π m}{q B} (1 - \frac{θ}{π})$